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# Fund Signals ECE 3500

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This 26 page Class Notes was uploaded by Shyanne Lubowitz on Monday October 26, 2015. The Class Notes belongs to ECE 3500 at University of Utah taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/230014/ece-3500-university-of-utah in ELECTRICAL AND COMPUTER ENGINEERING at University of Utah.

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ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 1 Chapter 6 Continuous Time System Realization Using the Laplace Transform In this chapter we start with a short review of Laplace transform and develop the notion of transfer functions of linear time invariant continuoustime LTIC systems Transfer functions are important because Any LTIC system can be characterized by a transfer function Transfer functions are useful mathematical tools for proposing alternate realizations of systems Transfer functions may also be used in evaluating the outputs of systems when their inputs are given 39 Given some specifications one may find a transfer function that satisfies the specifications This can then be used to realize the desired system In this chapter we take note of the first three points The last point will be addressed in a later chapter T fli l of this Chapter related to marshals in pages Bail 3924 and J l l 4 1f Latin gt1 l39gt4N39gtl ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 2 61 The Laplace Transform From your previous classes you may remember the definition of Laplace transform and its inverse for a signal fl as M mm f fte39 dt 61 Where s is a complex number and Where c is a constant chosen so that over the integration path ie s varying from C j00 to 6 13900 Fs exists In other words the path of integration in 62 must be within the region of convergence or existence of F s The shorthand for reference to 61 and 62 is fl gtFS ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 3 Example 61 Find the Laplace transform of f t e39 ut 3 Where MO is the Step function Solution 00 00 1 00 F S e ale stdl e saldl e sat J J S a 0 To evaluate the last term we let saoc j and note that sal sal 9 0 1 However 9 Hm can eX1st only when sal 0 a ReS a gt 0 In that case 9 Hm We thus obtaln l Fs Resa gt0 S a or equot ul ltgt Res gt a RES gt a is the region of convergence RC of Fs Fig 61 depicts the RC of F s 7 Resgta I Q Fig 61 Example 62 Find the Laplace transform of f t emu t ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 4 Answer Res lt a equot u l ltgt 77f Reslta N Fig 62 We note that muf and Fatwa have the same Laplace transform but different RC ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 5 Equations 61 and 62 are known as the bilateral Laplace transform pair or the twosided Laplace transform pair In the rest of this class to simplify our discussion we assume that all signals are causal ie are nonzero for positive t In that case RC is always on the righthand side of a vertical line such as S a and thus there will be no confusion as in the above pair of examples When t is nonzero only for t 2 0 61 will become M paw dz 63 The corresponding inverse Laplace transform will remain as in 62 This is known as Unilateral Laplace transform So in the rest of this class we use unilateral Laplace transform without mentioning the term unilateral The use of 039 is to make sure that in our derivations we include the Laplace transform of 6t correctlyunambiguously Exercise 61 Find the Laplace transform of a 51 b W ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 6 Finding the inverse Laplace transform Finding the inverse Laplace transforms involves integration in complex plane In this class we find inverse Laplace transform using table look up method Table 61 in page 372 of Lathi s book contains a comprehensive list of Laplace transforms of the most useful signals Note that all the timedomain functions in Table 61 are causal Example 63 Find the inverse Laplace transform of 7s 6 2 5 6Q3 sz S 6 0 s23s2 Q salnm3 Solution a To be able to use Table 61 we proceed with the following expansion 7S 6 7s 6 Flts2 s s 6 s2s 3 k1 k2 S2 s 3 k1 and k2 can be obtained as follows 7s 6 392 s 3 k2 Fss 3 i3 3 The above procedure is know as Heaviside method Thus F 4 3 S2 S 3 hFQMD and using Table 61 we get ft 4e ZZul 3e3 ut 4 3e3 ul ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 7 b We note that 2 Fs22S 52 k1 k2 s 3S2 s1 s2 2 7 13 S1 s2 where k1 and k2 are obtained as in Part 3 Using Table 61 we obtain ft 25r 7e l3e3921ut 0 Here 6 s 34 F l Ss 10s 34 6s 34 k1 k2 k ss5 j3s5j3 s s5 j3 s5j3 Using the method of Part a we obtain h6 mdk23j45ynw Finally using Table 61 of Lathi s book pair 10b we obtain fz 6 lOe395t cos z1269 uz ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform Example 64 Find the inverse Laplace transforms of 5s 2 3s2 s 5 3 a 52 5s6 b S2S 12 c S29 ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 9 Properties of Laplace transform We only review the relevant properties of the Laplace transform to our discussion in this chapter More properties of the Laplace transform can be found in Lathi s book Here we assume that t is causal ie is zero for I lt 0 1 Linearity Laplace transform is a linear transformation That is if f1f ltgt F1S andf2l gt F2S then k1f1lk2f2l ltgt k1F1S k2F2S Example 65 From Table 61 of the text we find that the Laplace transform of f quot011 is Use this result to find the Laplace transforms of a 1 2tut b 1 3t2ut c 1 t2ut ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 10 2 Time shifting If f0 ltgt FS then for tzo fTTo ltgt FSe Example 66 Find the Laplace transform of the function ft depicted in Fig 63 f V 2 4 Fig 63 Solution First we note that ft tut 2t 2ut 2 t 4ut 4 From Table 61 and the timeshifting property we have 141 ltgt 2 S 2v z 2ut 2 gt s2 e 4s s2 2 4ut 4 gt Thus 26 3925 63945 S2 Fs1 ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 11 Example 67 Find the inverse Laplace transform of e Zs a sl sl3e3923 b s 2s 3 c 1l 46 26 2S s ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 12 3 The timedifferentiation property If f I ltgt F S then lt sFs f039 and in general dquot n n n n drip ltgtS FSS S f 10 When f0 is causal the above results simplify to i sFs dz and d f ltgt squotFs dzquot Example 68 Noting that for the function fl given in Fig 63 we have the results of Fig 64 Use these results and the timedifferentiation and timeshifting properties of Laplace transform to obtain F s df A dt 1 7 4 I 1 V Fig 64 dzf alt2 ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 13 Solution 2 itz6t 26t 26t 4 ltgt s2Fs1 2e e394s Hence Ze39zs e39 2 S Fs1 4 The timeintegration property If f I ltgt F S then f frdr lt3 0 s Example 69 1 Note that MTd7 W0 and the Laplace transform of ut E Using these results and the time integration property find the Laplace transform of WC 5 Scaling If f I ltgt F S then for a gt 0 s fat ltgt a a ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 14 62 Transfer Functions We may recall from the earlier lectures in this class that the input output relationship of many LTIC systems has the following form n n l m m l ad ya d ya0ytbmdfb d f b t n din n l dtn1 m l dtm1 Taking Laplace transform on both sides of this equation and assuming that t 0 for I lt0 and the system is at rest all the initial conditions are zero we obtain ansquotYs an1squot391Ys a0Ys bmsmFs bm1squotquot1Fs b0Fs 0139 as as 391a Y s b smb s 391b Fs 1 1 11 0 m ml 0 Rearranging this result we obtain Ys bms quot bm1s quot391 130 Fs ansquot an1squot391 a0 This shows that the ratio of the Laplace transform of the system output to the Laplace transform of the system input is given by bsmb sm391b m m l 0 HS n n1 ans an1s aO The function H s is called the transfer function of the system We may also note that ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 15 YS FSHS This suggests a method of calculating the system output if its transfer function is known i obtain the Laplace transform of the input F s ii obtain the Laplace transform of output as YS FSHS iii obtain the output as W L391Ys Example 610 Solve the secondorder differential equation 12 51 6yt D lft where D and y039 140 0 and f0 41qu Solution The equation is dzy dy df 5 6 t t dt2 dt yo dt f0 We can look at this as a system with input t output yl and the transfer function s 1 H s s2 5s 6 On the other hand FSs4 ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 16 Thus sl YSFSHSW sl s2s3s4 l2 2 32 s2 s3 s4 and yt L391Ys e39z 2e393 gem ua Example 611 Solve the secondorder differential equation of example 610 when the initial conditions are y039 2 and 0 1 Solution Here we need to recall that dquot n n n n dzf lt2 s Flts s 1f0 s 70 flt W0 Replacing t by yt for n l and 2 we get dy EltgtsYs y0 sYs 2 and dzy gt s2Ys s 0 0 2Y 2 1 W y y S S S Substituting these in the equation d 2y dy df 5 6 t t dig dt y dtf we will nd an equation from which we can obtain Y S ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 17 63 Block Diagram Implementation of a large system generally involves dividing the system in to subsystems each of a complexity that can be handled Direct realization of a system with a large highorder transfer function is also not advised Such realization will be very sensitive to component tolerances We thus introduce the following realizations 1 Direct realization It is used only when the system has a low order less than or equal to 3 gtF Y 2 Cascade realization If HS H1 SH2 S the following realization is possible L W Y ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 18 3 Parallel realization If H S H1S H 2 S the following realization is possible Fs Y S 4 Feedback realization F S Y S gt ms Gsgt Fs l HsGs ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 19 64 System Realization We introduce the concept of system realization from transfer functions This is done through a simple example Fig 65 depicts a third order transfer function and a cascade realization of that Fs bs3bs2bsb Ys 33 22 l 0 39 s a2s alsa0 F s l X Y gr S l73s3l72s2blsb0 r s a2s alsa0 Fig 65 We first note that l 3 2 s a2s 611s61O Xs Fs This may be rearranged as s3Xs a2s2Xs alsXs aOXs Fs From this result we obtain the block diagram of Fig 66 as a system which satisfies this equation ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 20 Fig 66 On the other hand Ys b3s3Xs b2s2Xs blsXs b0Xs Noting that the terms S3XSS2XSSXS and X S are already available in various points of Fig 66 we obtain the complete block diagram of a realization of the desired transfer function as in Fig 67 In Fig 67 we have given an sdomain frequencydomain realization of the system In time domain the blocks US are replaced by integrators this is obvious from the time integration property of the Laplace transform Thus the timedomain equivalent of Fig 67 is the one shown in Fig 68 The structure of Figs 67 and 68 are known as canonical realization The term canonica refers to the fact that there are minimum number of integrators used in the structures ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 21 Fs s2Xs l sXs l Xs m YS gt 9 ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform Example 612 Find the canonical realization of the following transfer functions 1 s 2 5s 2s 3 a b c d s1 s1 s23s1 s25s6 22 ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 23 65 Use of Operational Amplifiers in System Realization The structure developed in the previous section uses adders integrators and amplifiers to realize linear systems You have learnt to make such elements in your previous classes We re introduced them here again for our purpose R R f0 ft yt o M I 39 39 k Rx f Fig 69 Op amp inverting amplifier and its symbol C I R f t yt f t e yt I a ytkfft L RC Fig 610 Op amp integrator and its symbol ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 24 R AAfA V V V R1 R 39 20 39 mt r s z yt mt frl ya k1fitquot39krfrt R R I k1f krf r R1 Rr Fig 611 Op amp summer with adjustable gain for different inputs Example 613 Using op amp circuits suggest a realization of the transfer function 2s 5 H S 2 s 4s 10 Solution Fig 612 depicts a block diagram of the canonical realization of H YS Fig 612 ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform 25 In converting the canonical realization to block diagram consist of summers and integrators we note that the polarity of the signals changes when go through each of these stages Taking this into account from Fig 612 we obtain the following block diagram F s Ys sXs l Xs Using this diagram we can obtain the op amp circuit diagram of the desired system ECE 3500 Chapter 6 CT Syst Analysis Using the Laplace Transform Example 614 In example 613 we assumed that unit of the poles and zeros of Hs were radians Assuming that unit of the poles and zeros of Hs be in kilo radians amend the nal circuit diagram Answer Divide all capacitors by 1000 Example 615 Suggest a circuit diagram for realization of the transfer function s2 2s 2 s2 2s 2 Assume that units of poles and zeros of Hs are in radians Hs Example 616 Suggest a cascade circuit diagram for realization of the transfer function 1 HS s ls2 Es 1 Assume that units of poles and zeros of Hs are in kilo radians 26

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